Junction Transistor Questions in English

(B) In a common base configuration, the input current is the emitter current $(I_e)$ and the output current is the collector current $(I_c)$.
The current gain $(\alpha)$ is defined as the ratio of the change in collector current to the change in emitter current: $\alpha = \frac{\Delta I_c}{\Delta I_e}$.
Since $I_e = I_c + I_b$ and the base current $(I_b)$ is always positive, $I_c < I_e$, which implies $\alpha < 1$.
Thus, the Assertion is correct.
The collector-base junction is indeed reverse-biased in a transistor to ensure that the majority of charge carriers injected from the emitter are collected at the collector, which is essential for amplification.
Thus, the Reason is also correct.
However, the fact that the collector is reverse-biased explains why the transistor works as an amplifier, but it is not the direct reason why the current gain is less than unity (which is due to the definition of the common base configuration where $I_c < I_e$).
Therefore, both are correct, but the Reason is not the correct explanation of the Assertion.

(A) The input impedance of a common emitter configuration is defined as $Z_{in} = \left| \frac{\Delta V_{BE}}{\Delta i_B} \right|_{V_{CE} = \text{constant}}$.
In a common emitter transistor, the base-emitter junction is forward-biased to allow current flow.
Because the base-emitter junction is forward-biased, the base current $\Delta i_B$ is relatively large for a small change in base-emitter voltage $\Delta V_{BE}$.
Since $Z_{in} = \frac{\Delta V_{BE}}{\Delta i_B}$, a small change in voltage divided by a relatively large change in current results in a low input impedance.
Therefore, both the Assertion and the Reason are correct, and the Reason is the correct explanation for the Assertion.

(B) The transistor is considered the basic constitutional unit of modern electronic circuits. It acts as a fundamental building block for integrated circuits (ICs),microprocessors,and memory devices,functioning as either an electronic switch or an amplifier.

(N/A) No,the law of conservation of energy is not violated. The increase in the energy of the output $AC$ signal is derived from the energy stored in the $DC$ power supply (battery).
When an amplifier is used,the $DC$ source provides the necessary energy to boost the weak input signal. Therefore,the energy of the output signal is equal to the energy of the input $AC$ signal plus the energy drawn from the $DC$ supply.
Thus,the law of conservation of energy is strictly maintained.

(A-D) In the given common emitter transistor circuit, the input circuit consists of the base-emitter junction with resistance $R_1$.
When $R_1$ is increased, the base current $I_B$ decreases according to Ohm's law $(I_B = V_{in} / R_1)$.
The output collector current $I_C$ is related to the base current by the relation $I_C = \beta I_B$, where $\beta$ is the current gain of the transistor.
Since $I_B$ decreases, the collector current $I_C$ also decreases.
The ammeter is in series with the collector, so its reading decreases.
The voltmeter is connected across the resistor $R_2$ in the output circuit. The voltage across $R_2$ is given by $V_{out} = I_C R_2$.
As $I_C$ decreases, the voltage drop $V_{out}$ across $R_2$ decreases.
Therefore, the readings of both the ammeter and the voltmeter will decrease.

(D) In the input section:
$V_i = I_B R_B + V_{BE}$
Given $V_i = 10\, V$,$R_B = 400\, k\Omega = 400 \times 10^3\, \Omega$,and $V_{BE} = 0\, V$.
$10 = I_B (400 \times 10^3) + 0$
$I_B = \frac{10}{400 \times 10^3} = 0.025 \times 10^{-3}\, A = 25 \times 10^{-6}\, A = 25\, \mu A$
In the output section:
$V_{CC} = I_C R_C + V_{CE}$
Given $V_{CC} = 10\, V$,$R_C = 3\, k\Omega = 3 \times 10^3\, \Omega$,and $V_{CE} = 0\, V$.
$10 = I_C (3 \times 10^3) + 0$
$I_C = \frac{10}{3 \times 10^3} = 3.33 \times 10^{-3}\, A = 3.33\, mA = 3333\, \mu A$
Calculating current gain $\beta$:
$\beta = \frac{I_C}{I_B} = \frac{3.33 \times 10^{-3}}{25 \times 10^{-6}} = \frac{3333}{25} = 133.32$

(N/A) From the graph in figure $(2)$, for the operating point $Q$, we have $I_B = 30 \, \mu A$, $I_C = 4 \, mA = 4 \times 10^{-3} \, A$, and $V_{CE} = 8 \, V$.
For the output circuit:
$V_{CC} = I_C R_C + V_{CE}$
$16 = (4 \times 10^{-3}) R_C + 8$
$R_C = \frac{8}{4 \times 10^{-3}} = 2 \times 10^3 \, \Omega = 2 \, k\Omega$.
For the input circuit:
$V_{BB} = I_B R_B + V_{BE}$
$16 = (30 \times 10^{-6}) R_B + 0.7$
$R_B = \frac{15.3}{30 \times 10^{-6}} = 0.51 \times 10^6 \, \Omega = 510 \, k\Omega$.
Voltage gain $A_V = \beta \frac{R_C}{R_{in}}$. Assuming input resistance $R_{in} \approx \frac{V_{BE}}{I_B} = \frac{0.7}{30 \times 10^{-6}} \approx 23.33 \, k\Omega$.
$A_V = \left( \frac{I_C}{I_B} \right) \left( \frac{R_C}{R_{in}} \right) = \left( \frac{4 \times 10^{-3}}{30 \times 10^{-6}} \right) \left( \frac{2000}{23333} \right) \approx 133.33 \times 0.0857 \approx 11.43$.
Power gain $A_P = A_V \times \beta = 11.43 \times 133.33 \approx 1524$.

(C) The $AC$ current gain $\beta_{ac}$ is defined as the ratio of the change in collector current $\Delta I_{C}$ to the change in base current $\Delta I_{B}$ at a constant collector-emitter voltage $V_{CE}$.
From the graph,at $V_{CE} = 10\, V$,the collector current $I_{C} = 4.0\, mA$ corresponds to a base current $I_{B} = 30\, \mu A$.
To find $\beta_{ac}$ around this operating point,we consider the adjacent curves for $I_{B} = 20\, \mu A$ and $I_{B} = 40\, \mu A$.
At $V_{CE} = 10\, V$:
For $I_{B} = 20\, \mu A$,$I_{C} = 3.0\, mA$.
For $I_{B} = 40\, \mu A$,$I_{C} = 6.0\, mA$.
Therefore,the change in base current is $\Delta I_{B} = 40\, \mu A - 20\, \mu A = 20\, \mu A = 20 \times 10^{-6}\, A$.
The change in collector current is $\Delta I_{C} = 6.0\, mA - 3.0\, mA = 3.0\, mA = 3.0 \times 10^{-3}\, A$.
The $AC$ current gain is $\beta_{ac} = \frac{\Delta I_{C}}{\Delta I_{B}} = \frac{3.0 \times 10^{-3}}{20 \times 10^{-6}} = \frac{3000}{20} = 150$.

(A) For effective transistor action,the base region must be very thin and lightly doped to ensure that most of the charge carriers injected from the emitter reach the collector without recombining in the base.

(D) The voltage drop across the load resistance $R_C$ is given by Ohm's law as:
$V_L = I_C \times R_C$
Given that the voltage drop $V_L = 0.8 \; V$ and the load resistance $R_C = 800 \; \Omega$.
Substituting these values into the equation:
$0.8 \; V = I_C \times 800 \; \Omega$
$I_C = \frac{0.8}{800} \; A$
$I_C = 0.001 \; A$
Since $1 \; A = 1000 \; mA$,we have:
$I_C = 0.001 \times 1000 \; mA = 1 \; mA$.
Therefore,the collector current is $1 \; mA$.

(C) In the saturation region, the collector-emitter voltage $V_{CE}$ is approximately $0 \, V$.
From the output loop circuit:
$V_{CC} = I_C R_{out} + V_{CE}$
Since $V_{CE} = 0 \, V$ at saturation:
$3 = I_C (200)$
$I_C = \frac{3}{200} = 0.015 \, A = 15 \, mA$
Using the relation $\beta = \frac{I_C}{I_B}$, we find the base current $I_B$ required for saturation:
$I_B = \frac{I_C}{\beta} = \frac{15 \times 10^{-3}}{200} = 75 \times 10^{-6} \, A = 75 \, \mu A$
Now, applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:
$V_{BB} = I_B R_{in} + V_{BE}$
$V_{BB} = (75 \times 10^{-6} \, A)(100 \times 10^3 \, \Omega) + 0.7 \, V$
$V_{BB} = 7.5 \, V + 0.7 \, V = 8.2 \, V$
Thus, the required value of $V_{BB}$ is $8.2 \, V$.

(B) The voltage gain $A_v$ of a common emitter amplifier is defined as the ratio of the output voltage to the input voltage.
It is calculated using the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Given:
Input resistance $R_i = 3 \, \Omega$
Load resistance $R_L = 24 \, \Omega$
Current gain $\beta = 0.6$
Substituting the values into the formula:
$A_v = 0.6 \times \frac{24}{3}$
$A_v = 0.6 \times 8$
$A_v = 4.8$
Therefore,the voltage gain is $4.8$.

(B) The power gain $(A_p)$ of a common emitter amplifier is given by the formula: $A_p = \beta^2 \times \frac{R_o}{R_i}$, where $\beta$ is the current gain, $R_o$ is the output resistance, and $R_i$ is the input resistance.
Given: $A_p = 10^6$, $R_i = 100\, \Omega$, and $R_o = 10\, k\Omega = 10^4\, \Omega$.
Substituting the values into the formula:
$10^6 = \beta^2 \times \frac{10^4}{10^2}$
$10^6 = \beta^2 \times 10^2$
$\beta^2 = \frac{10^6}{10^2} = 10^4$
$\beta = \sqrt{10^4} = 100$.
Thus, the common emitter current gain $\beta$ is $100$.

(D) The current gain $\alpha$ is defined as the ratio of collector current $(I_{C})$ to emitter current $(I_{E})$: $\alpha = \frac{I_{C}}{I_{E}}$.
The current gain $\beta$ is defined as the ratio of collector current $(I_{C})$ to base current $(I_{B})$: $\beta = \frac{I_{C}}{I_{B}}$.
From Kirchhoff's current law for a transistor,the emitter current is the sum of base and collector currents: $I_{E} = I_{B} + I_{C}$.
Substituting $I_{E}$ in the expression for $\alpha$: $\alpha = \frac{I_{C}}{I_{B} + I_{C}}$.
Dividing the numerator and denominator by $I_{C}$: $\alpha = \frac{1}{\frac{I_{B}}{I_{C}} + 1}$.
Since $\frac{1}{\beta} = \frac{I_{B}}{I_{C}}$,we substitute this into the equation: $\alpha = \frac{1}{\frac{1}{\beta} + 1} = \frac{1}{\frac{1+\beta}{\beta}} = \frac{\beta}{1+\beta}$.
Therefore,the correct relation is $\alpha = \frac{\beta}{1+\beta}$.

(A) The current gain $\beta$ for a common-emitter configuration is defined as the ratio of the change in collector current $\Delta I_C$ to the change in base current $\Delta I_B$ at a constant collector-emitter voltage $V_{CE}$.
$\beta = \frac{\Delta I_C}{\Delta I_B}$
From the given graph,let us consider the change between two curves,for example,from $I_B = 10 \ \mu A$ to $I_B = 20 \ \mu A$ in the active region.
At $I_B = 10 \ \mu A$,the collector current $I_C = 2 \ mA = 2 \times 10^{-3} \ A$.
At $I_B = 20 \ \mu A$,the collector current $I_C = 4 \ mA = 4 \times 10^{-3} \ A$.
Therefore,$\Delta I_C = (4 - 2) \ mA = 2 \ mA = 2 \times 10^{-3} \ A$.
And $\Delta I_B = (20 - 10) \ \mu A = 10 \ \mu A = 10 \times 10^{-6} \ A$.
Now,calculating the current gain:
$\beta = \frac{2 \times 10^{-3}}{10 \times 10^{-6}} = \frac{2}{10} \times 10^3 = 0.2 \times 1000 = 200$.
Thus,the estimated current gain is $200$.

(A) The relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is given by $I_E = I_C + I_B$.
For small changes,this can be written as $\Delta I_E = \Delta I_C + \Delta I_B$.
Given $\Delta I_E = 4\, mA$ and $\Delta I_C = 3.5\, mA$,we find the change in base current:
$\Delta I_B = \Delta I_E - \Delta I_C = 4\, mA - 3.5\, mA = 0.5\, mA$.
The current gain $\beta$ is defined as the ratio of change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3.5\, mA}{0.5\, mA} = 7$.
Therefore,the value of $\beta$ is $7$.

(A) Statement $I$ is false because a transistor is a single crystal device with specific doping profiles and regions (emitter,base,collector). Connecting two $PN$ junction diodes back-to-back does not create a functional transistor because it lacks the necessary charge carrier injection and control mechanisms inherent in a monolithic transistor structure.
Statement $II$ is true. The current amplification factor $\beta$ (or $h_{fe}$) for a common-emitter configuration is defined as the ratio of the collector current $(i_c)$ to the base current $(i_b)$,expressed as $\beta = \frac{i_c}{i_b}$.

(D) The active region of the $CE$ transistor is the linear region where the output current is directly proportional to the input current. This linearity makes it the best-suited region for operating the transistor as an amplifier.

(B) Given: $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$.
We know that the emitter current is the sum of collector current and base current: $I_{E} = I_{C} + I_{B}$.
Substituting $I_{E}$ in the expression for $\alpha$:
$\alpha = \frac{I_{C}}{I_{C} + I_{B}}$.
Dividing the numerator and denominator by $I_{B}$:
$\alpha = \frac{I_{C}/I_{B}}{(I_{C}/I_{B}) + (I_{B}/I_{B})} = \frac{\beta}{\beta + 1}$.
Now,rearrange to solve for $\beta$:
$\alpha(\beta + 1) = \beta$
$\alpha \beta + \alpha = \beta$
$\alpha = \beta - \alpha \beta$
$\alpha = \beta(1 - \alpha)$
$\beta = \frac{\alpha}{1 - \alpha}$.

(C) Given:
Collector resistance $R_C = 1000 \, \Omega$
Voltage drop across collector resistor $\Delta V_C = 0.6 \, V$
Current gain $\beta = 24$
The collector current $I_C$ is given by Ohm's law:
$I_C = \frac{\Delta V_C}{R_C} = \frac{0.6 \, V}{1000 \, \Omega} = 0.6 \times 10^{-3} \, A = 6 \times 10^{-4} \, A$
The relationship between collector current $I_C$ and base current $I_B$ is:
$I_C = \beta \times I_B$
Therefore,the base current $I_B$ is:
$I_B = \frac{I_C}{\beta} = \frac{6 \times 10^{-4} \, A}{24} = 0.25 \times 10^{-4} \, A$
Converting to microamperes $(\mu A)$:
$I_B = 0.25 \times 10^{-4} \times 10^6 \, \mu A = 25 \, \mu A$.

(B) The input resistance $r_i$ is calculated as the ratio of the change in base-emitter voltage to the change in base current:
$r_i = \frac{\Delta V_{BE}}{\Delta I_B} = \frac{10 \, mV}{10 \, \mu A} = \frac{10 \times 10^{-3} \, V}{10 \times 10^{-6} \, A} = 1000 \, \Omega = 1 \, k\Omega$.
The current gain $\beta$ is the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{1.5 \, mA}{10 \, \mu A} = \frac{1.5 \times 10^{-3} \, A}{10 \times 10^{-6} \, A} = 150$.
The voltage gain $A_V$ for a common-emitter amplifier is given by the product of the current gain and the ratio of load resistance $R_L$ to input resistance $r_i$:
$A_V = \beta \times \frac{R_L}{r_i} = 150 \times \frac{5000 \, \Omega}{1000 \, \Omega} = 150 \times 5 = 750$.

(D) The voltage gain $A_v$ is given by the product of current gain $\beta$ and resistance gain.
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{(15 - 5) \times 10^{-3} \text{ A}}{(300 - 100) \times 10^{-6} \text{ A}} = \frac{10 \times 10^{-3}}{200 \times 10^{-6}} = 50$.
The voltage gain $A_v = \beta \times \frac{R_o}{R_i} = 50 \times \frac{60 \ \Omega}{200 \ \Omega} = 50 \times 0.3 = 15$.

(D) transistor acts as a switch by toggling between two states: the $Cut-off$ state and the $Saturation$ state.
In the $Cut-off$ state,the transistor is $OFF$ (no current flows).
In the $Saturation$ state,the transistor is $ON$ (maximum current flows).
Therefore,to function as a switch,it must be operated in both the $Saturation$ and $Cut-off$ states.

(A) In a transistor,the current is carried by charge carriers. In an $n-p-n$ transistor,the majority charge carriers are electrons,whereas in a $p-n-p$ transistor,the majority charge carriers are holes.
Electrons have higher mobility compared to holes because they are lighter and interact less with the crystal lattice.
Due to this higher mobility,electrons can move faster through the base region,allowing for higher current conduction in $n-p-n$ transistors compared to $p-n-p$ transistors.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(A) Given: Change in base current $\Delta i_{B} = 100 \,\mu A = 100 \times 10^{-6} \,A = 0.1 \,mA$.
Change in collector current $\Delta i_{C} = 10 \,mA$.
Load resistance $R_{L} = 2 \,k \Omega = 2000 \,\Omega$.
Input resistance $R_{in} = 1 \,k \Omega = 1000 \,\Omega$.
First, calculate the current gain $\beta$:
$\beta = \frac{\Delta i_{C}}{\Delta i_{B}} = \frac{10 \,mA}{0.1 \,mA} = 100$.
Power gain $A_{P}$ is given by the formula:
$A_{P} = \beta^{2} \times \frac{R_{L}}{R_{in}}$.
Substituting the values:
$A_{P} = (100)^{2} \times \frac{2000 \,\Omega}{1000 \,\Omega} = 10000 \times 2 = 20000$.
$A_{P} = 2 \times 10^{4}$.
Comparing this with $x \times 10^{4}$, we get $x = 2$.

(B) The current gain $\beta$ in $CE$ configuration is given by $\beta = \frac{\Delta I_C}{\Delta I_B}$.
From the graph,we can choose two points to calculate the change in collector current $\Delta I_C$ and base current $\Delta I_B$.
Let us take the points $(I_B = 100\,\mu A, I_C = 5\,mA)$ and $(I_B = 200\,\mu A, I_C = 10\,mA)$.
$\Delta I_C = (10 - 5)\,mA = 5\,mA = 5 \times 10^{-3}\,A$.
$\Delta I_B = (200 - 100)\,\mu A = 100\,\mu A = 100 \times 10^{-6}\,A$.
Therefore,$\beta = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = \frac{5000}{100} = 50$.
The voltage gain $A_V$ is given by $A_V = \beta \times \frac{R_C}{R_{in}}$.
Given $R_C = 2\,k\Omega$ and $R_{in} = 0.50\,k\Omega$.
$A_V = 50 \times \frac{2\,k\Omega}{0.50\,k\Omega} = 50 \times 4 = 200$.

(B) The small signal current gain (current amplification factor) $\beta_{ac}$ is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage $(V_{CE})$.
Given:
Change in collector current,$\Delta I_C = 6\,mA - 4\,mA = 2\,mA = 2 \times 10^{-3}\,A$
Change in base current,$\Delta I_B = 25\,\mu A - 20\,\mu A = 5\,\mu A = 5 \times 10^{-6}\,A$
The formula for current gain is:
$\beta_{ac} = \frac{\Delta I_C}{\Delta I_B}$
Substituting the values:
$\beta_{ac} = \frac{2 \times 10^{-3}}{5 \times 10^{-6}}$
$\beta_{ac} = \frac{2}{5} \times 10^3$
$\beta_{ac} = 0.4 \times 1000 = 400$
Thus,the current amplification factor is $400$.

(B) The voltage gain $A_v$ of a common emitter amplifier is given by the formula:
$A_v = \frac{v_{\text{out}}}{v_{\text{in}}} = \beta \frac{R_{\text{out}}}{R_{\text{in}}}$
Given:
$\beta = 100$
$R_{\text{in}} = 1 \text{ k}\Omega = 10^3 \, \Omega$
$R_{\text{out}} = 10 \text{ k}\Omega = 10^4 \, \Omega$
$v_{\text{in}} = 1 \text{ mV} = 10^{-3} \, \text{V}$
Substituting the values:
$v_{\text{out}} = v_{\text{in}} \times \beta \times \frac{R_{\text{out}}}{R_{\text{in}}}$
$v_{\text{out}} = 10^{-3} \times 100 \times \frac{10 \times 10^3}{1 \times 10^3}$
$v_{\text{out}} = 10^{-3} \times 100 \times 10$
$v_{\text{out}} = 1 \, \text{V}$

(B) The power gain of a transistor is defined as the product of current gain and voltage gain.
In a Common Emitter $(CE)$ configuration, both the current gain $(\beta)$ and the voltage gain are significantly high.
In a Common Base $(CB)$ configuration, the current gain $(\alpha)$ is less than $1$, resulting in low power gain.
In a Common Collector $(CC)$ configuration, the voltage gain is less than $1$, resulting in lower power gain compared to the $CE$ configuration.
Therefore, the power gain is highest in the Common Emitter configuration.

(C) In a transistor,the base is made very thin and lightly doped to ensure that the majority of charge carriers injected from the emitter into the base pass through to the collector.
If the base were heavily doped,the number of majority charge carriers in the base would be high.
This would lead to a higher rate of recombination between the injected charge carriers from the emitter and the majority charge carriers in the base.
By keeping the base lightly doped,the recombination rate is significantly decreased,which allows most of the charge carriers to reach the collector,thereby increasing the current gain of the transistor.

(D) The current gain $\beta$ in a common emitter $(CE)$ configuration is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.
Given:
Change in base current,$\Delta I_B = 200 \ \mu A - 100 \ \mu A = 100 \ \mu A = 100 \times 10^{-6} \ A$.
Change in collector current,$\Delta I_C = 16.5 \ mA - 9 \ mA = 7.5 \ mA = 7.5 \times 10^{-3} \ A$.
Using the formula:
$\beta = \frac{\Delta I_C}{\Delta I_B}$
Substituting the values:
$\beta = \frac{7.5 \times 10^{-3} \ A}{100 \times 10^{-6} \ A}$
$\beta = \frac{7.5 \times 10^{-3}}{10^{-4}}$
$\beta = 7.5 \times 10^{1} = 75$.
Therefore,the value of current gain $\beta$ is $75$.

(D) The voltage gain $A_v$ of a common emitter amplifier is given as $1000$.
In a common emitter amplifier,there is a phase shift of $\pi$ radians $(180^{\circ})$ between the input and output signals.
The input signal is $v_i = (0.004 \, V) \sin(\omega t + \pi/2)$.
The output signal amplitude $V_0$ is given by $V_0 = A_v \times V_i = 1000 \times 0.004 \, V = 4 \, V$.
Due to the phase shift of $\pi$,the output signal is $v_o = 4 \sin(\omega t + \pi/2 + \pi) = 4 \sin(\omega t + 3\pi/2)$.
Since $\sin(\theta + 3\pi/2) = -\cos(\theta) = \sin(\theta - \pi/2)$,the output signal is $v_o = 4 \sin(\omega t - \pi/2) \, V$.

(C) In a common base $(CB)$ configuration,the current gain $\alpha$ is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in emitter current $(\Delta I_E)$.
$\alpha = \frac{\Delta I_C}{\Delta I_E}$
Given:
$\alpha = 0.98$
$\Delta I_E = 5.00 \,mA$
To find the change in collector current $(\Delta I_C)$:
$\Delta I_C = \alpha \times \Delta I_E$
$\Delta I_C = 0.98 \times 5.00 \,mA$
$\Delta I_C = 4.90 \,mA$
Therefore,the change in collector current is $4.9 \,mA$.

(D) The power gain $(A_p)$ of an amplifier is defined as the product of voltage gain $(A_v)$ and current gain $(A_i)$.
Mathematically,$A_p = A_v \times A_i$.
Given:
Power gain $(A_p)$ = $7.5$
Voltage gain $(A_v)$ = $2.5$
Substituting the values into the formula:
$7.5 = 2.5 \times A_i$
$A_i = \frac{7.5}{2.5} = 3$.
Therefore,the current gain is $3$.

(A) Given:
Input resistance $R_i = 2 \, k\Omega = 2 \times 10^3 \, \Omega$.
$A$.$C$. current gain $\beta = 20$.
Load resistance $R_L = 5 \, k\Omega$.
The transconductance $g_m$ of a transistor is defined as the ratio of the output current change to the input voltage change,which is given by the formula:
$g_m = \frac{\beta}{R_i}$
Substituting the given values:
$g_m = \frac{20}{2 \times 10^3 \, \Omega}$
$g_m = 10 \times 10^{-3} \, \Omega^{-1}$
$g_m = 0.01 \, \Omega^{-1}$
Thus,the transconductance is $0.01 \, \Omega^{-1}$.

(C) The relationship between the emitter current $(\Delta I_E)$,collector current $(\Delta I_C)$,and base current $(\Delta I_B)$ in a transistor is given by the equation:
$\Delta I_E = \Delta I_C + \Delta I_B$
Given:
$\Delta I_E = 7.89 \, mA$
$\Delta I_C = 7.8 \, mA$
Substituting the values into the equation:
$7.89 \, mA = 7.8 \, mA + \Delta I_B$
$\Delta I_B = 7.89 \, mA - 7.8 \, mA$
$\Delta I_B = 0.09 \, mA$
Therefore,the change in base current required is $0.09 \, mA$.

(B) Given:
$Current \, amplification \, factor \, (\beta) = 60$
$Input \, resistance \, (R_i) = 1 \, k\Omega = 1000 \, \Omega$
The transconductance $(g_m)$ of a transistor is defined as the ratio of the current amplification factor to the input resistance:
$g_m = \frac{\beta}{R_i}$
Substituting the given values:
$g_m = \frac{60}{1000} \, \Omega^{-1}$
$g_m = 0.06 \, S = 6 \times 10^{-2} \, S$
Thus, the transconductance is $6 \times 10^{-2} \, SI \, units$.

(A) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Where $\beta$ is the current gain,$R_L$ is the load resistance,and $R_i$ is the input resistance.
Given:
$\beta = 50$
$R_L = 9 \, k\Omega = 9000 \, \Omega$
$R_i = 500 \, \Omega$
Substituting the values into the formula:
$A_v = 50 \times \frac{9000}{500}$
$A_v = 50 \times 18$
$A_v = 900$
Therefore,the voltage gain of the amplifier is $900$.

(D) The correct answer is $D$.
$A$ transistor is a three-terminal semiconductor device used for amplification and switching. It can be configured as an amplifier,an oscillator,or a modulator.
$A$ rectifier is a device that converts alternating current $(AC)$ to direct current $(DC)$. Rectification is typically performed using $P-N$ junction diodes,not transistors. Therefore,a transistor cannot be used as a rectifier.

(B) The power gain $(A_p)$ of a common emitter $(CE)$ amplifier is given by the formula:
$A_p = \beta^2 \times \frac{R_L}{R_i}$
Where:
$\beta = 6$ (current gain)
$R_L = 24 \, k\Omega$ (load resistance)
$R_i = 3 \, k\Omega$ (input resistance)
Substituting the values into the formula:
$A_p = (6)^2 \times \left( \frac{24 \, k\Omega}{3 \, k\Omega} \right)$
$A_p = 36 \times 8$
$A_p = 288$
Thus,the power gain is $288$.

(B) Given:
Common base current gain $\alpha = 0.99$
Load resistance $R_L = 4.5 \, k\Omega = 4500 \, \Omega$
Dynamic input resistance $R_i = 50 \, \Omega$
The voltage gain $A_v$ for a common base amplifier is given by the formula:
$A_v = \alpha \cdot \frac{R_L}{R_i}$
Substituting the given values:
$A_v = 0.99 \times \frac{4500}{50}$
$A_v = 0.99 \times 90$
$A_v = 89.1$
Thus,the voltage gain of the amplifier is $89.1$.

(B) Given: $\beta = 50$,$I_B = 40 \, \mu A = 40 \times 10^{-6} \, A$,$V_{CC} = 10 \, V$,$R_C = 2 \, k\Omega = 2000 \, \Omega$.
Step $1$: Calculate the collector current $I_C$.
$I_C = \beta \times I_B = 50 \times 40 \times 10^{-6} \, A = 2000 \times 10^{-6} \, A = 2 \times 10^{-3} \, A = 2 \, mA$.
Step $2$: Apply Kirchhoff's Voltage Law to the output loop.
$V_{CE} = V_{CC} - I_C R_C$.
Step $3$: Substitute the values.
$V_{CE} = 10 \, V - (2 \times 10^{-3} \, A) \times (2000 \, \Omega) = 10 \, V - 4 \, V = 6 \, V$.
Therefore,the collector-emitter voltage $V_{CE}$ is $6 \, V$.

(B) In the given circuit diagram,the emitter terminal of the transistor is directly connected to the ground.
Since the emitter terminal is common to both the input (base-emitter) and output (collector-emitter) circuits,this configuration is known as the common emitter configuration.

(C) When amplifiers are connected in series,the total voltage gain $(A_v)$ is the product of the individual voltage gains of each amplifier.
Given that there are $3$ amplifiers,each with a voltage gain of $A_1 = 10$,$A_2 = 10$,and $A_3 = 10$.
The resultant gain $A_{total} = A_1 \times A_2 \times A_3$.
$A_{total} = 10 \times 10 \times 10 = 1000$.
Therefore,the resultant gain is $1000$.

(A) Given: Current gain $\beta = 50$,Load resistance $R_C = 5 \, k\Omega = 5000 \, \Omega$,Voltage across load $V_C = 5 \, V$.
First,calculate the collector current $I_C$ using Ohm's law:
$I_C = \frac{V_C}{R_C} = \frac{5 \, V}{5000 \, \Omega} = 1 \times 10^{-3} \, A = 1 \, mA$.
Now,use the relation for current gain $\beta$ to find the base current $I_B$:
$\beta = \frac{I_C}{I_B} \implies I_B = \frac{I_C}{\beta}$.
Substituting the values:
$I_B = \frac{1 \, mA}{50} = 0.02 \, mA$.
Therefore,the base current is $0.02 \, mA$.

(A) The output $ac$ voltage is $V_{out} = 2.0\,V$. The collector resistance is $R_C = 2.0\,k\Omega = 2000\,\Omega$.
The $ac$ collector current is $i_C = V_{out} / R_C = 2.0\,V / 2000\,\Omega = 1.0\,mA$.
The signal base current is $i_B = i_C / \beta = 1.0\,mA / 100 = 0.010\,mA$.
The $dc$ base current $I_B$ is given as $10$ times the signal current: $I_B = 10 \times 0.010\,mA = 0.10\,mA$.
Using the base circuit equation $V_{BB} = I_B R_B + V_{BE}$,we assume the standard base-emitter voltage $V_{BE} = 0.6\,V$.
$R_B = (V_{BB} - V_{BE}) / I_B = (2.0\,V - 0.6\,V) / 0.10\,mA = 1.4\,V / 0.10\,mA = 14\,k\Omega$.

(B) Given:
Current gain $(\beta)$ = $62$
Collector resistance $(R_C)$ = $5\,k\Omega = 5000\,\Omega$
Input resistance $(R_{in})$ = $500\,\Omega$
Input voltage $(V_{in})$ = $0.01\,V$
The voltage gain $(A_v)$ of a common emitter amplifier is given by:
$A_v = \beta \times \frac{R_C}{R_{in}}$
$A_v = 62 \times \frac{5000}{500} = 62 \times 10 = 620$
The output voltage $(V_o)$ is given by:
$V_o = A_v \times V_{in}$
$V_o = 620 \times 0.01\,V = 6.2\,V$
Therefore,the output voltage is $6.2\,V$.

(A) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the formula: $A_v = \beta \times \frac{R_L}{R_{in}}$, where $\beta$ is the current gain, $R_L$ is the load resistance, and $R_{in}$ is the input resistance.
Given: $\beta = 50$, $R_L = 5 \, k\Omega = 5000 \, \Omega$, and input peak voltage $V_{in} = 5 \, mV = 0.005 \, V$.
Assuming the input resistance $R_{in}$ is $1 \, k\Omega$ (as implied by the ratio in the provided context), the voltage gain is calculated as:
$A_v = 50 \times \frac{5 \, k\Omega}{1 \, k\Omega} = 50 \times 5 = 250$.